To determine absolute convergence, we check whether the series $\sum |a_n|$ converges.

Series (i):

$a_n = \frac{(-1)^n}{n^2}$ The absolute value is: $\left| \frac{(-1)^n}{n^2} \right| = \frac{1}{n^2}$ This is a p-series of the form $\sum \frac{1}{n^p}$, where $p = 2$. Since $p > 1$, the series converges absolutely.

Series (ii):

$a_n = \frac{(-1)^n}{n (\ln n)^3}$ The absolute value is: $\left| \frac{(-1)^n}{n (\ln n)^3} \right| = \frac{1}{n (\ln n)^3}$ We apply the integral test to determine convergence of: $\int_{2}^{\infty} \frac{dx}{x (\ln x)^3}$ Using substitution $u = \ln x$, so that $du = \frac{dx}{x}$, the integral transforms into: $\int \frac{du}{u^3}$ which evaluates to: $\frac{-1}{2u^2} = \frac{-1}{2 (\ln x)^2}$ As $x \to \infty$, $\frac{-1}{2 (\ln x)^2}$ approaches 0, meaning the integral converges. Hence, the series $\sum \frac{1}{n (\ln n)^3}$ converges absolutely.

Final Answer:

Both series (i) and (ii) converge absolutely.

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Here are 5 related questions you might find useful:

1. What is the difference between absolute and conditional convergence?
2. How do you apply the ratio test to determine absolute convergence?
3. What is the p-series test, and why does it work?
4. How does the integral test help in determining convergence?
5. Can a series converge conditionally but not absolutely?

Tip: A series that converges absolutely also converges, but a series that converges conditionally may still diverge absolutely!